On the other hand for t b a 2 E t E E t So x u tx t u tx and hence u tx u x for

# On the other hand for t b a 2 e t e e t so x u tx t u

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On the other hand, for 0 t ( b - a )/ 2 , 0 E ( t ) E ( 0 ) = 0 E ( t ) = 0 . So x u ( t,x ) = t u ( t,x ) 0 and hence u ( t,x ) = u ( 0 ,x ) = 0 for a + t x b - t , that is, u 0 in = {( x,t ) ( -∞ , ) × [ 0 , ) a + t x b - t } . Food for Thought. Please read the Tutorial 9 Problem 1. Problem 2. (i) Let u 1 and u 2 be two solution of the problem. Define ˜ u = u 1 - u 2 . Then we have t ˜ u - 2 xx ˜ u = - ˜ u for 0 < x < L, t > 0 ˜ u t = 0 = ˜ u x = 0 = x ˜ u x = L = 0 (2) Multiplying the PDE t ˜ u = 2 xx ˜ u - ˜ u by ˜ u , and then integrating with respect to x over ( 0 ,L ) , we have L 0 ˜ u∂ t ˜ udx = 2 L 0 ˜ u∂ xx ˜ udx - L 0 ˜ u 2 dx 2

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MATH4406 Introduction to PDE 1 2 d dt L 0 ˜ u 2 dx = 2 [ ˜ u∂ x ˜ u ] L x = 0 - 2 L 0 x ˜ u 2 dx - L 0 ˜ u 2 dx = - 2 L 0 x ˜ u 2 dx - L 0 ˜ u 2 dx (by (2)) 0 . Thus, L 0 ˜ u ( t,x )∣ 2 dx L 0 ˜ u ( 0 ,x )∣ 2 dx = 0 (by (2)) and hence ˜ u = 0 , that is, u 1 = u 2 . (ii) Let u 1 and u 2 be two solution of the problem. Define ˜ u = u 1 - u 2 . Then we have tt ˜ u - 9 xx ˜ u = - ˜ u - 2 t ˜ u for 0 < x < L, t > 0 ˜ u t = 0 = t ˜ u t = 0 = x ˜
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